contestada

Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conductor, with respect to the outer conductor, is +600 V.
A) An electron is released from rest at the surface of the outer conductor. The speed of the electron as it reaches the inner conductor is closest to:__________.
B) The maximum electric field magnitude between the cylinders is closest to:_______.

Respuesta :

Answer:

a) The speed of the electron as it reaches the inner conductor is closest to:

v = 1.45 × 10⁷m/s

b) The electric field magnitude between the cylinders is

E = 10,000V/m

Explanation:

given

inner radius of the cylinder r₁ = 20mm = 0.02m

outter radius of the cylinder r₂ = 80mm = 0.08m

potential difference V= 600V

mass of electron = 9.1×10⁻³¹kg

charge on electron = 1.6×10⁻¹⁹C

calculating the work done in bringing electron at inner conductor is

[tex]W =\frac{1}{2}mv^{2}[/tex]

note:

[tex]V = \frac{W}{q}[/tex]

∴W = (ΔV)q

(ΔV)q = [tex]\frac{1}{2}mv^{2}[/tex]

(600)1.6×10⁻¹⁹ = ¹/₂ × 9.1×10⁻³¹ × v²

v² ≈ 2.11 × 10¹⁴

v = 1.45 × 10⁷m/s

According to the energy conservation law, the total energy of an isolated system is always constant.  

The energy of an isolated system can neither be created nor be destroyed, it can only convert one form to another form.

∴ the maximum electric field

E = ΔV/d

E = 600/d

where d is the distance between the two points

where d = 0.06m

E = 600/0.06

E = 10,000V/m

Note: the electric field due to the potential difference between to points depends upon the potential difference V and the distance between both points d.

a) The speed of the electron as it reaches the inner conductor is closest to: v = 1.45 × 10⁷m/s

b) The electric field magnitude between the cylinders is, E = 10,000V/m

Given:

Inner radius of the cylinder r₁ = 20mm = 0.02m

Outer radius of the cylinder r₂ = 80mm = 0.08m

Potential difference V= 600V

Mass of electron = [tex]9.1*10^{-31}kg[/tex]

Charge on electron = 1.6×10⁻¹⁹C

A)

Calculation for Work Done:

[tex]W=1/2mv^2[/tex]............(1)

Also.

[tex]V=\frac{W}{q}[/tex]

Thus, [tex]W=\triangle V*q[/tex]...........(2)

On equating 1 and 2:

[tex]\triangle V*q=1/2mv^2\\\\(600)1.6*10^{-19} = 1/2 * 9.1*10^{-31}* v^2\\\\v^2 =2.11 * 10^{14}\\\\v = 1.45 * 1067m/s[/tex]

B)

Law of conservation of Energy:

The energy of an isolated system can neither be created nor be destroyed, it can only convert one form to another form.

Thus, the maximum electric field

[tex]E = \triangle V/d\\\\E = 600/d[/tex]

where d is the distance between the two points

d = 0.06m

[tex]E = 600/0.06\\\\E = 10,000V/m[/tex]

Thus, the maximum electric field is 10,000V/m.

Find more information about Electric Field here:

brainly.com/question/14372859

Otras preguntas